Integrating factor mistake?

Question

Can someone help me get the Integrating factor for this? $$y'\tan x+ y =e^{2x}\sin x$$

I got tan^sec²X But this was actually et multiple choice question and my teacher told me to retry the question as this was not amongst the options.

Answer 1

$$y'\tan x + y =e^{2x}\sin x$$ Multiply by $\mu(x)= \cos x$ $$\implies y'\sin x + y \cos x =e^{2x}\sin x \cos x$$ $$\implies (y\sin x)' =\frac 12e^{2x}\sin(2x)$$

Integrate ...

Answer 2

Get your integral into the from

$y' + P(x) y = Q(x)$

and your integrating factor is $e^{\int P(x) \ dx}$

and then:

$e^{\int P(x) \ dx}y' + P(t)e^{\int P(x) \ dt} y = Q(x)e^{\int P(x) \ dx} \\ e^{\int P(x) \ dx} y = Q(x)\int e^{\int P(x) \ dx} \ dx $

In this case.

$y' + (\cot x) y = Q(x)\\ \int \cot x\ dx = \ln \sin x\\ e^{\ln\sin x} = \sin x$

But as Isham points out, it is not entirely necessary to divide through by $\tan x$ find in the integrating factor of $\sin x$ and multiply through by the integrating factor if you realise that $\int (\sin x) y' + (\cos x) y \ dx = (\sin x) y$

Answer 3

Well, the integrating factor technique works as follows. We multiply the equation with $f(x)$ such that the left-hand side becomes $(f(x) y(x))'$. In this case, we multiply with $f/\tan$ to make our life easier. Then $f$ solves $$ f'(x) = f(x)/\tan(x). $$ Bringing $f$ to the otherside and integrating we get $$ \log(f(x)) = \int \frac{1}{\tan(x)}\,\mathrm{d} x = \log(\sin(x)). $$ Hence, the integrating factor will be $$ \sin(x)/\tan(x) = \cos(x). $$

Answer 4

$$y'+\cot(x)y=e^{2x}\cos(x)\\ \mu(x)=e^{\int \cot(x)dx}$$

$$\mu=e^{\ln\sin(x)} \\ \mu=\sin(x)$$

$$y\sin(x)=\frac12 \int e^{2x}\sin(2x)dx$$

$$y\sin(x)=\frac{1}{32} e^{2x}[\sin(2x)-4\cos(2x)]$$

$$\color{red}{y(x)= \frac{1}{32} e^{2x}[1-4\cot(2x)]}$$